Area between a parabola and a line

Intersections. $x^2=x+2\iff x^2-x-2=0\iff(x-2)(x+1)=0$, so $x=-1$ or $x=2$.

Which is on top. At $x=0$ the line has height $2$ and the parabola $0$, so the line is above on $[-1,2]$.

Area.
$$\int_{-1}^2\left(x+2-x^2\right)dx=\left[\frac{x^2}2+2x-\frac{x^3}3\right]_{-1}^2=\left(2+4-\frac83\right)-\left(\frac12-2+\frac13\right)=\frac{10}3+\frac76=\frac92.$$

Answer. The area is $\dfrac92$.